我正在努力學習Spring.我使用以下工具使用Spring Boot創(chuàng)建了一個項目：

> Spring Data JPA
> Spring Data REST
>春天的HATEOAS
>春季安全

我正在嘗試創(chuàng)建一個用戶實體.我希望用戶擁有加密密碼(鹽).

當我對/ api / users進行POST時,我成功創(chuàng)建了一個新用戶.

{
"firstname":"John",
"lastname":"Doe",
"email":"johndoe@example.com",
"password":"12345678"
}

但我有兩個問題：

>密碼以明文形式保存
>鹽是空的

06001

我認為問題是使用默認構(gòu)造函數(shù)而不是我創(chuàng)建的另一個.我是Spring和JPA的新手,所以我必須遺漏一些東西.這是我的代碼.

User.java

@Entity
@Table(name = "users")
public class User{

    @Id
    @GeneratedValue
    private Long id;

    @Column(nullable = false)
    public String firstname;

    @Column(nullable = false)
    public String lastname;

    @Column(nullable = false, unique = true)
    public String email;

    @JsonIgnore
    @Column(nullable = false)
    public String password;

    @JsonIgnore
    @Column
    private String salt;

    public User() {}

    public User(String email, String firstname, String lastname, String password) {
        this.email = email;
        this.firstname = firstname;
        this.lastname = lastname;
        this.salt = UUID.randomUUID().toString();
        this.password = new BCryptPasswordEncoder().encode(password   this.salt);
    }


    @JsonIgnore
    public String getSalt() {
        return salt;
    }

    @JsonProperty
    public void setSalt(String salt) {
        this.salt = salt;
    }

    public String getEmail() {
        return email;
    }

    public void setEmail(String email) {
        this.email = email;
    }

    public String getFirstname() {
        return firstname;
    }

    public void setFirstname(String firstname) {
        this.firstname = firstname;
    }

    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    public String getLastname() {
        return lastname;
    }

    public void setLastname(String lastname) {
        this.lastname = lastname;
    }

    @JsonIgnore
    public String getPassword() {
        return password;
    }

    @JsonProperty
    public void setPassword(String password) {
        this.password = password;
    }

}

UserRepository.java

public interface UserRepository extends JpaRepository<User, Long> {

    public User findByEmail(String email);

    public User findByEmailAndPassword(String email, String password);
}

Application.java

@SpringBootApplication
public class Application {


    public static void main(String[] args) {
        SpringApplication.run(Application .class, args);
    }

}

此外,如果有人發(fā)現(xiàn)我做錯了什么,我想指出我應(yīng)該把用戶登錄代碼放在哪里/如何(解密).

謝謝.

解決方法:

所以,這就是我如何解決我的問題：我創(chuàng)建了一個Controller作為我的自定義端點,然后我創(chuàng)建了一個服務(wù),在其中我放置了我想要創(chuàng)建用戶的邏輯.這是代碼：

UserController.java

@Controller
public class UserController {

    @Autowired
    private UserService userService;

    @RequestMapping("/api/register")
    @ResponseBody
    public Long register(@RequestBody User user) {
        return userService.registerUser(user);
    }

    ...

}

UserService .java

@Service
public class UserService {

    @Autowired
    private UserRepository userRepository;

    public Long registerUser(User user) {
        user.setPassword(new BCryptPasswordEncoder().encode(password));
        userRepository.save(user);
        return user.getId();
    }

        ...

}

所以通過POST來做

{
"firstname":"John",
"lastname":"Doe",
"email":"johndoe@example.com",
"password":"12345678"
}

在/ api / register中,我現(xiàn)在可以創(chuàng)建一個帶有哈希密碼的用戶.